Proving that the composition of relations are equal to each other Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When is the composition of partial orders a partial order?About binary relations under certain conditions and their compositionComposite Relations - Give Examples of relations $R_1$ and $R_2$ such that $R_2 circ R_1 = R_1 circ R_2$ and $R_2 circ R_1 neq R_1 circ R_2$Suppose that R and S are reflexive relations on a set A. Prove or disprove each of these statements.Suppose R and S are relations $subseteq Atimes B$Let $Q$ be the symmetric closure of $R$, and let $S$ be the transitive closure of $Q$. Prove that $S$ is the symmetric transitive closure of $R$.Notation for the composition of functions and relations.Symmetricity of composition of equivalence relationsWhy equality of relations is defined like that?Correct definition of composite relations?
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Proving that the composition of relations are equal to each other
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When is the composition of partial orders a partial order?About binary relations under certain conditions and their compositionComposite Relations - Give Examples of relations $R_1$ and $R_2$ such that $R_2 circ R_1 = R_1 circ R_2$ and $R_2 circ R_1 neq R_1 circ R_2$Suppose that R and S are reflexive relations on a set A. Prove or disprove each of these statements.Suppose R and S are relations $subseteq Atimes B$Let $Q$ be the symmetric closure of $R$, and let $S$ be the transitive closure of $Q$. Prove that $S$ is the symmetric transitive closure of $R$.Notation for the composition of functions and relations.Symmetricity of composition of equivalence relationsWhy equality of relations is defined like that?Correct definition of composite relations?
$begingroup$
Please excuse my English if it's not understandable, exercises are translated so I don't know all the English terms in math.
So I'm doing exercises regarding relations and compositions, and one of the exercises is:
Suppose $Rsubseteq Atimes B$, $Ssubseteq Btimes C$ and $Tsubseteq Ctimes D$. Which of the following statements are true?
- $(Scirc R)^-1=R^-1circ S^-1$
- $Scirc R=R^-1circ S^-1$
- $Scirc R=(Rcirc S)^-1$
- $(Tcirc S)circ R=Tcirc (Scirc R)$
The answer to 1.:
To prove 1., suppose $(c,a)in(Scirc R)^-1$. Then, by definition of the inverse relation, $(a,c)in (Scirc R)$. By the definition of the composite relation, there exists $bin B$ such that $(a,b)in R$ and $(b,c)in S$. This of course means $(c,b)in S^-1$ and $(b,a)in R^-1$. Applying the definition of the composite relation again, we find $(c,a)in R^-1circ S^-1$. Thus, $(Scirc R)subseteq R^-1circ S^-1$. Similar arguments show $R^-1circ S^-1subseteq (Scirc R)$ and the result follows.
Note that this is not my answer. I just don't understand how I can come to this conclusion. Our lecturer hasn't covered how to prove these kind of calcualtions.
Can someone please explain to me the approach to these kind of exercises?
discrete-mathematics relations
edited Mar 26 at 18:44
st.math
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
$endgroup$
add a comment |
$begingroup$
Please excuse my English if it's not understandable, exercises are translated so I don't know all the English terms in math.
So I'm doing exercises regarding relations and compositions, and one of the exercises is:
Suppose $Rsubseteq Atimes B$, $Ssubseteq Btimes C$ and $Tsubseteq Ctimes D$. Which of the following statements are true?
- $(Scirc R)^-1=R^-1circ S^-1$
- $Scirc R=R^-1circ S^-1$
- $Scirc R=(Rcirc S)^-1$
- $(Tcirc S)circ R=Tcirc (Scirc R)$
The answer to 1.:
To prove 1., suppose $(c,a)in(Scirc R)^-1$. Then, by definition of the inverse relation, $(a,c)in (Scirc R)$. By the definition of the composite relation, there exists $bin B$ such that $(a,b)in R$ and $(b,c)in S$. This of course means $(c,b)in S^-1$ and $(b,a)in R^-1$. Applying the definition of the composite relation again, we find $(c,a)in R^-1circ S^-1$. Thus, $(Scirc R)subseteq R^-1circ S^-1$. Similar arguments show $R^-1circ S^-1subseteq (Scirc R)$ and the result follows.
Note that this is not my answer. I just don't understand how I can come to this conclusion. Our lecturer hasn't covered how to prove these kind of calcualtions.
Can someone please explain to me the approach to these kind of exercises?
discrete-mathematics relations
edited Mar 26 at 18:44
st.math
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
$endgroup$
add a comment |
$begingroup$
Please excuse my English if it's not understandable, exercises are translated so I don't know all the English terms in math.
So I'm doing exercises regarding relations and compositions, and one of the exercises is:
Suppose $Rsubseteq Atimes B$, $Ssubseteq Btimes C$ and $Tsubseteq Ctimes D$. Which of the following statements are true?
- $(Scirc R)^-1=R^-1circ S^-1$
- $Scirc R=R^-1circ S^-1$
- $Scirc R=(Rcirc S)^-1$
- $(Tcirc S)circ R=Tcirc (Scirc R)$
The answer to 1.:
To prove 1., suppose $(c,a)in(Scirc R)^-1$. Then, by definition of the inverse relation, $(a,c)in (Scirc R)$. By the definition of the composite relation, there exists $bin B$ such that $(a,b)in R$ and $(b,c)in S$. This of course means $(c,b)in S^-1$ and $(b,a)in R^-1$. Applying the definition of the composite relation again, we find $(c,a)in R^-1circ S^-1$. Thus, $(Scirc R)subseteq R^-1circ S^-1$. Similar arguments show $R^-1circ S^-1subseteq (Scirc R)$ and the result follows.
Note that this is not my answer. I just don't understand how I can come to this conclusion. Our lecturer hasn't covered how to prove these kind of calcualtions.
Can someone please explain to me the approach to these kind of exercises?
discrete-mathematics relations
edited Mar 26 at 18:44
st.math
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
$endgroup$
Please excuse my English if it's not understandable, exercises are translated so I don't know all the English terms in math.
So I'm doing exercises regarding relations and compositions, and one of the exercises is:
Suppose $Rsubseteq Atimes B$, $Ssubseteq Btimes C$ and $Tsubseteq Ctimes D$. Which of the following statements are true?
- $(Scirc R)^-1=R^-1circ S^-1$
- $Scirc R=R^-1circ S^-1$
- $Scirc R=(Rcirc S)^-1$
- $(Tcirc S)circ R=Tcirc (Scirc R)$
The answer to 1.:
To prove 1., suppose $(c,a)in(Scirc R)^-1$. Then, by definition of the inverse relation, $(a,c)in (Scirc R)$. By the definition of the composite relation, there exists $bin B$ such that $(a,b)in R$ and $(b,c)in S$. This of course means $(c,b)in S^-1$ and $(b,a)in R^-1$. Applying the definition of the composite relation again, we find $(c,a)in R^-1circ S^-1$. Thus, $(Scirc R)subseteq R^-1circ S^-1$. Similar arguments show $R^-1circ S^-1subseteq (Scirc R)$ and the result follows.
Note that this is not my answer. I just don't understand how I can come to this conclusion. Our lecturer hasn't covered how to prove these kind of calcualtions.
Can someone please explain to me the approach to these kind of exercises?
discrete-mathematics relations
discrete-mathematics relations
edited Mar 26 at 18:44
st.math
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
edited Mar 26 at 18:44
st.math
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
edited Mar 26 at 18:44
st.math
1,268115
edited Mar 26 at 18:44
st.math
1,268115
edited Mar 26 at 18:44
st.math
1,268115
1,268115
asked Mar 26 at 17:53
bendikbpbendikbp
62
asked Mar 26 at 17:53
bendikbpbendikbp
62
asked Mar 26 at 17:53
bendikbpbendikbp
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $Asubseteq B$ and $Bsubseteq A$. Now, if you want to show that $Asubseteq B$, you have to prove that the implication $$xin Aimplies xin B$$ is true (similar for $Bsubseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)in(Scirc R)^-1$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)in R^-1circ S^-1$ is also true.
The basic properties to use here are:
- $(a,b)in Siff(b,a)in S^-1$
- $(a,c)in Scirc Riffexists_bin B,(a,b)in Rland (b,c)in S$
answered Mar 26 at 17:58
st.mathst.math
1,268115
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $Asubseteq B$ and $Bsubseteq A$. Now, if you want to show that $Asubseteq B$, you have to prove that the implication $$xin Aimplies xin B$$ is true (similar for $Bsubseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)in(Scirc R)^-1$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)in R^-1circ S^-1$ is also true.
The basic properties to use here are:
- $(a,b)in Siff(b,a)in S^-1$
- $(a,c)in Scirc Riffexists_bin B,(a,b)in Rland (b,c)in S$
answered Mar 26 at 17:58
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $Asubseteq B$ and $Bsubseteq A$. Now, if you want to show that $Asubseteq B$, you have to prove that the implication $$xin Aimplies xin B$$ is true (similar for $Bsubseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)in(Scirc R)^-1$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)in R^-1circ S^-1$ is also true.
The basic properties to use here are:
- $(a,b)in Siff(b,a)in S^-1$
- $(a,c)in Scirc Riffexists_bin B,(a,b)in Rland (b,c)in S$
answered Mar 26 at 17:58
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $Asubseteq B$ and $Bsubseteq A$. Now, if you want to show that $Asubseteq B$, you have to prove that the implication $$xin Aimplies xin B$$ is true (similar for $Bsubseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)in(Scirc R)^-1$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)in R^-1circ S^-1$ is also true.
The basic properties to use here are:
- $(a,b)in Siff(b,a)in S^-1$
- $(a,c)in Scirc Riffexists_bin B,(a,b)in Rland (b,c)in S$
answered Mar 26 at 17:58
st.mathst.math
1,268115
$endgroup$
Note that to show that two sets are equal (relations are sets), e.g. $A=B$, then it has to be true that $Asubseteq B$ and $Bsubseteq A$. Now, if you want to show that $Asubseteq B$, you have to prove that the implication $$xin Aimplies xin B$$ is true (similar for $Bsubseteq A$).
This is what the solution of the first question does. It takes an arbitrary element $(c,a)in(Scirc R)^-1$ and then concludes by properties of (inverse) relations and by using the definition of relations, that $(c,a)in R^-1circ S^-1$ is also true.
The basic properties to use here are:
- $(a,b)in Siff(b,a)in S^-1$
- $(a,c)in Scirc Riffexists_bin B,(a,b)in Rland (b,c)in S$
answered Mar 26 at 17:58
st.mathst.math
1,268115
edited Mar 26 at 18:05
answered Mar 26 at 17:58
st.mathst.math
1,268115
answered Mar 26 at 17:58
st.mathst.math
1,268115
answered Mar 26 at 17:58
st.mathst.math
1,268115
1,268115
add a comment |
add a comment |
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